Calculus BC ยท Unit 2
A nearby secant slope approximates the instantaneous rate of change. Use the closest table or graph points.
Jump to section:
Instantaneous rate
The instantaneous rate of change of \(g\) at \(a\) is the slope of the tangent. A difference quotient with nearby points approximates that slope.
The quotient
\(\dfrac{g(a + h) - g(a)}{h}\) or \(\dfrac{g(b) - g(a)}{b - a}\) with \(b\) as close to \(a\) as the table or graph allows.
Closest points win
On a table, use the x-values nearest the target. On a graph, use the nearest lattice points you can read.
Write it, do not have to evaluate
If the question says write a difference quotient, the expression is the answer. You do not have to simplify it to a single number unless you are asked.
Write the difference quotient from a table.
| \(x\) | 1.8 | 1.9 | 1.99 | 2 | 2.01 | 2.1 | 2.2 |
|---|---|---|---|---|---|---|---|
| \(g(x)\) | 2.40 | 2.71 | 2.97 | 3.00 | 3.04 | 3.32 | 3.65 |
The closest table values to 2 are 1.99 and 2.01, or 2 and 2.01.
Using 2 and the nearest right-hand value:
\(\dfrac{g(2.01) - g(2)}{2.01 - 2}\)
Answer: \(\dfrac{g(2.01) - g(2)}{0.01}\)
| \(x\) | 1.8 | 1.9 | 1.99 | 2 | 2.01 | 2.1 | 2.2 |
|---|---|---|---|---|---|---|---|
| \(g(x)\) | 2.40 | 2.71 | 2.97 | 3.00 | 3.04 | 3.32 | 3.65 |
\(\dfrac{g(2) - g(1.99)}{2 - 1.99}\)
Answer: \(\dfrac{g(2) - g(1.99)}{0.01}\)
| \(x\) | 1.8 | 1.9 | 1.99 | 2 | 2.01 | 2.1 | 2.2 |
|---|---|---|---|---|---|---|---|
| \(g(x)\) | 2.40 | 2.71 | 2.97 | 3.00 | 3.04 | 3.32 | 3.65 |
\(\dfrac{g(2.01) - g(1.99)}{2.01 - 1.99}\)
Answer: \(\dfrac{g(2.01) - g(1.99)}{0.02}\)
\(\dfrac{h(4.1) - h(4)}{4.1 - 4}\)
Answer: \(\dfrac{h(4.1) - h(4)}{0.1}\)
\(\dfrac{p(3.05) - p(3)}{0.05}\)
Answer: \(\dfrac{p(3.05) - p(3)}{0.05}\)
Farther x-values give a secant that is less like the tangent.
\(1.8\) and \(2.2\) is worse.
Answer: \(1.8\) and \(2.2\)
Read nearby points from a graph.
Graph of g
The nearest lattice points around 2 are \((1, 3)\) and \((3, 2)\), or 2 with a neighbor.
Using \(x = 2\) and \(x = 3\):
\(\dfrac{g(3) - g(2)}{3 - 2}\)
From the graph, \(g(2) = 4\) and \(g(3) = 2\).
\(\dfrac{2 - 4}{1} = -2\) is the numerical slope if a number is asked. The quotient itself is the requested expression.
Answer: \(\dfrac{g(3) - g(2)}{1}\)
Graph of g
Nearby points \((0, 1)\) and \((2, 4)\), or 1 with a neighbor.
\(\dfrac{g(2) - g(1)}{2 - 1}\)
Answer: \(\dfrac{g(2) - g(1)}{1}\)
Graph of g
Nearby: \((3, 2)\) and \((5, 1)\).
\(\dfrac{g(5) - g(3)}{5 - 3} = \dfrac{1 - 2}{2}\)
\(-\dfrac{1}{2}\)
Answer: \(-\dfrac{1}{2}\)
Graph of g
The graph is rising through \(x = 1\).
Positive.
Answer: positive
Graph of g
The graph is falling through \(x = 3\).
Negative.
Answer: negative
Graph of g
\(\dfrac{g(1) - g(0)}{1 - 0}\)
Answer: \(\dfrac{g(1) - g(0)}{1}\)
A table that also shows a hole, or a two-sided estimate.
| \(x\) | 0.9 | 0.99 | 0.999 | 1 | 1.001 | 1.01 | 1.1 |
|---|---|---|---|---|---|---|---|
| \(f(x)\) | 1.82 | 1.94 | 1.97 | undef. | 2.03 | 2.07 | 2.21 |
\(f\) is undefined at 1, and the nearby y-values approach about 2 from both sides.
Removable discontinuity.
A reasonable estimate is the average of the closest values.
\(\dfrac{1.97 + 2.03}{2} = 2.00\)
Answer: removable; about 2.00
| \(x\) | 0.9 | 0.99 | 0.999 | 1 | 1.001 | 1.01 | 1.1 |
|---|---|---|---|---|---|---|---|
| \(f(x)\) | 1.82 | 1.94 | 1.97 | undef. | 2.03 | 2.07 | 2.21 |
Use the closest defined values on each side.
\(\dfrac{f(1.001) - f(0.999)}{1.001 - 0.999}\)
Answer: \(\dfrac{f(1.001) - f(0.999)}{0.002}\)
\(\dfrac{2.03 - 1.97}{0.002}\)
\(\dfrac{0.06}{0.002}\)
\(30\)
Answer: \(30\)
Those x-values are farther from 1, so the secant is less like the tangent.
Answer: the points are farther from 1
\(\dfrac{s(5.02) - s(5)}{0.02}\)
Answer: \(\dfrac{s(5.02) - s(5)}{0.02}\)
\(\dfrac{g(2.6) - g(2.5)}{2.6 - 2.5}\)
Answer: \(\dfrac{g(2.6) - g(2.5)}{0.1}\)
Need help with difference quotients?
Book a Session with Amy